To find the de-Broglie wavelength of the electron in the Bohr orbit of a hydrogenic atom, we can use the formula:

λ = h / (mv)

where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J·s), m is the mass of the electron, and v is the velocity of the electron.

In the Bohr model, the velocity of the electron can be calculated using the formula:

v = (Z * e^2) / (4πε₀ * h * n)

where Z is the atomic number (1 for hydrogen), e is the elementary charge (1.602 x 10^-19 C), ε₀ is the vacuum permittivity (8.854 x 10^-12 C^2/(N·m^2)), h is Planck's constant, and n is the principal quantum number (which determines the Bohr orbit).

Given that the de-Broglie wavelength is equal to 1.5, we can set up the following equation:

1.5 = (h / (m * v))

Rearranging the equation to solve for v:

v = h / (1.5 * m)

Now, substituting the formula for v in the Bohr model:

v = (Z * e^2) / (4πε₀ * h * n)

h / (1.5 * m) = (Z * e^2) / (4πε₀ * h * n)

We know that Z (atomic number) is 1 for hydrogen, so:

h / (1.5 * m) = (e^2) / (4πε₀ * h * n)

Simplifying, we can cancel out the h terms:

1 / (1.5 * m) = (e^2) / (4πε₀ * n)

Now, we can substitute the values for the constants:

1 / (1.5 * m) = (1.602 x 10^-19 C)^2 / (4π * 8.854 x 10^-12 C^2/(N·m^2) * n)

Simplifying further:

1 / (1.5 * m) = (2.566 x 10^-38 C^2) / (3.535 x 10^-11 C^2/(N·m^2) * n)

Since the mass of the electron (m) is approximately 9.109 x 10^-31 kg, we can substitute this value:

1 / (1.5 * 9.109 x 10^-31 kg) = (2.566 x 10^-38 C^2) / (3.535 x 10^-11 C^2/(N·m^2) * n)

Simplifying further:

1 / (1.36635 x 10^-30 kg) = (2.566 x 10^-38 C^2) / (3.535 x 10^-11 C^2/(N·m^2) * n)

Now, solving for n:

n = (2.566 x 10^-38 C^2) / (3.535 x 10^-11 C^2/(N·m^2) * (1.36635 x 10^-30 kg))

n ≈ 1